Question

已知等差数列{an}中,a3a7=-16,a4+a6=0,求{an}前n项和sn.要详解答案,我采纳！急用！

Answer 1

a3a7=(a1+2d)(a1+6d)=a1^2+8a1d+12d^2=-16
a4+a6=a1+3d+a1+5d=2a1+8d=0 a1=-4d
代入得
(-4d)^2+8*(-4d)d+12d^2=-16
d=2或d=-2
a1=-8或a1=8
Sn=na1+n(n-1)d/2
sn=n^2-9n或 sn=-n^2+9n

Answer 2

a4+a6=0,所以a5=0,由a3a7=-16,所以（a5-2d)(a5+2d)=-16,把，d解出来，再根据a5,把通向解出来，再用公式把sn解出来，希望采纳

Answer 3

设公差为d
a4+a6=2a5=2(a1+4d)=0
a1=-4d
a3a7=(a1+2d)(a1+6d)=a1^2+8a1d+12d^2=-16
即-4d^2=-16 d=±2
(1) d=2时 a1=-8
an=-8+2(n-1)=2n-10
sn=n(2n-18)/2=n^2-9n
(2) d=-2时 a1=8
an=10-2n
sn=9n-n^2